warewolfe

Thanks for the reply and it was a help, but default.jpg and the target.jpg did have different names. (Which was the bit that weirded me out). Your code does make the browser distinguish between default.jpg and target.jpg but if I update target.jpg with the same form it takes a manual refresh for the page to notice the difference.  

Hej  = is an assignment operator, a variable is assigned a value, =& is a reference operator, two variables reference one value.  $varA = 1;  $varA now has the value of 1 $varA =& $varB  $varA now has the value of $varB, whatever that is.

Hej put "IndexIgnore *.gif *.jpg" in your htaccess file to block listing of image files. WW.

Hi, I have a same-page form-display script that calls a function to generate a jpg. If the form hasn't been filled in it shows a default .jpg. The trouble is that when the submit button is pushed and the new .jpg is created the page itself doesn't change to show the new image. Is there a php solution or do I have to use javascript? 

  Hi, Here is a tutorial that kind of explains how to do what you want to do. It also explains a lot more but you can always parse that out yourself. It has the "activate" function you seem to be after. http://www.phpfreaks.com/tutorials/65/1.php

Hej Humpty, Interesting problem you got there, A simple way would be if you had a form that took in the desired input on what the fonts and tables looks like and then wrote to a preexisting stylesheet and a page that linked to that css. First write a stylesheet stub and set the permissions so you can write to it prewrite all of the information that is static, then have the script append all the dynamic information. $file ="DynamicStyleSheet.css" $fh = fopen($file,'a') or die(echo 'File ($file) does not exist';) fputs($fh, $divContent); fclose($fh); (Quick and dirty, need much more comments/error handling here) where $divContent is a string containing the css code needed to generate the look for a certain div id or class. This should works but because it it doesn't overwrite the style sheet will get quite long after a while. I'll look into a better way later today. WW.

Hej, try <?php $link = mysql_connect('localhost', 'user', 'pass); if (!$link) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully<br>'; mysql_select_db("test",$link"); $result = mysql_query(SELECT * FROM people",$link); while($row = mysql_fetch_array($result,MYSQL_NUM)) { foreach($row as $val) { echo'<p>$val</p>'; } } ?>